# Full text of "Treatise On Analysis Vol-Ii"

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```6    NEGLIGIBLE FUNCTIONS AND SETS       119

oo

(13.6.1)    If(fn) is a sequence of negligible functions g:0, then £/„ is negligible.
This follows immediately from (13.5.8).

A subset N of X is said to be negligible (with respect to ju) or ^-negligible
if its characteristic function <pN (12.7) is /^-negligible. Clearly any subset of a
negligible set is negligible.

(13.6.2)    A denumerable union of negligible sets is negligible.

For if (Nfc) is a sequence of negligible sets and N = (J Nk , we have

k

= sup <pNk ^ £ cpNk, and the result follows from (13.6.1).

For example, with respect to Lebesgue measure A, every set {t} consisting
of a single real number is negligible. For if e > 0 there exists a function
/e «2f R(R) with values in [0, 1] which is equal to 1 at the point t and vanishes
on the complement of the interval [f — e, / + e], and therefore

A*(«p{r)) £ A(/) g 28.

Hence, by (13.6.2), it follows that every denumerable subset of R (for example,
the set Q of rational numbers) is negligible with respect to Lebesgue measure.
One can also give examples of nondenumerable sets in R which are
negligible with respect to Lebesgue measure (Section 13.8, Problem 4).

A property P(;t) of the points of X is said to be true almost everywhere
(with respect to IJL) if the complement of the set of points for which P(x) is
true is ju-negligible.

(13.6.3)   A mapping f: X-^R is negligible if and only if it is zero almost
everywhere.

Let N be the set of all jceX such that \f(x)\ > 0. Then we have
<pN g sup n\f\9 and |/| :g sup «<pN, hence the result follows from (13.5.7).

n                                             n

(1 3.6.4)   If a mapping f : X -» R is such that /^*(/) < + oo (resp. ju*(/) > - oo)>
thenf(x) < + oo (resp./(x) > — oo) almost everywhere.

It is enough to prove the assertion relating to the upper integral. By
hypothesis, there exists a function h e */ such that/^ h and /u*(A) < +00.
We may therefore limit ourselves to the case where /e »/, and since there is and form an
```